Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:

import java.util.Scanner;

public class Main {
    static double[][] aa;
    static double maxx(int h,int t,int n,double p[]){
        if(h+t==n &&  h>t  ){
            return (double)1;
        }
        if(h+t==n && t>h){
            return (double)0;
        }
        if(aa[h][t]!=-1){
            return aa[h][t];
        }
        return aa[h][t]= (double)((double)p[h+t]*(double)maxx(h+1,t,n,p)+(double)((double)1-(double)p[h+t])*(double)maxx(h,t+1,n,p)) ;
    }

    public static void main(String[] args) {
        Scanner scanner=new Scanner(System.in);
        int n=scanner.nextInt();
        double[] p=new double[n];
        for(int j=0;j<n;j++){
            p[j]= scanner.nextDouble();
        }
        aa=new double[n][n];
        for(int j=0;j<n;j++){
            for(int i=0;i<n;i++){
                aa[j][i]=-1;
            }
        }
        System.out.println(maxx(0,0,n,p));
    }


}